An Underdetermined Linear System for GPSDan Kalman

Dan Kalman (kalman@american.edu) joined themathematics faculty at American University in 1993,following an eight year stint in the aerospace industry andearlier teaching positions in Wisconsin and South Dakota.He has won three MAA writing awards and served a termas Associate Executive Director of the MAA. His interestsinclude matrix algebra, curriculum development, andinteractive computer environments for exploringmathematics, especially using Mathwright software.

Finding the general solution to an underdetermined linear system is a standard topicin linear algebra. It contributes to a complete analysis of the behaviors of linear sys-tems, as well as providing a foundation for understanding more abstract topics, includ-ing linear transformations, null space, and dimension. But at the first introduction ofthe topic it would be nice to have a simple, realistic example where the parameterizedgeneral solution of an underdetermined system is of practical interest. In this note, Iwill present such an example connected with the Global Positioning System (GPS) fordetermining geographical locations.

The basic idea of GPS is a variant on three dimensional triangulation: a point onthe surface of the earth is determined by its distances from three other points. Here,the point we wish to determine is the location of the GPS receiver, the other points aresatellites, and the distances are computed using the travel times of radio signals. Thisrequires accurate time keeping, prompting a slight modification of the pure spatial tri-angulation problem. In the modified version, we need four satellites, rather than three,and can then calculate both the location, and the correct time, at the GPS receiver.

Before presenting the example, I should make it clear that the computations that fol-low are not the same as the methods actually used by GPS. The example assumes exactgeometric knowledge, whereas GPS has to deal with real world measurement errors.Thus, GPS typically uses more than four satellites, and a least-squares method to deter-mine the best estimate of the location and time at the receiver. Other refinements in theactual GPS calculations take into account the way a radio signal is impeded by passingthrough the atmosphere, and the actual encoding of information in the radio signal. Foran accurate overview of how the GPS system actually works, see [2]. There is quite abit of linear algebra involved, and complete details are presented in [3]. Reference [4]provides an additional account of the mathematics involved in GPS computations.

Although the formulation I will present does not represent the methods used byGPS, it does provide a good starting point for understanding the GPS system. Forexample, the exact geometric model can be perturbed by random errors to simulate theeffects of errors in the real system. This idea will be considered briefly below.

The Geometric Model. For concreteness, consider a ship at sea in an unknownlocation. It has a GPS receiver that obtains simultaneous signals from four satellites.Each signal specifies its time of transmission and the position of the satellite at thattime. This allows the GPS receiver to compute its position and time. How does thiswork?

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To begin with, we imagine that there is an xyz-coordinate system with the earthcentered at the origin, the positive z axis running through the north pole and fixedrelative to the earth. The unknown position of the ship can be expressed as a point(x, y, z), which can later be translated into a latitude and longitude. To simplify things,let us mark off the three axes in units equal to the radius of the earth. Thus, a pointat sea level will have x2 + y2 + z2 = 1 in this system. Also, we will measure time inunits of milliseconds. The GPS system finds distances by knowing how long it takesa radio signal to get from one point to another. For this we need to know the speed oflight, approximately equal to .047 (in units of earth radii per millisecond).

Our ship is at an unknown position and has no clock. It receives simultaneous sig-nals from four satellites, giving their positions and times as shown in Table 1. (Thesenumbers were made up for the example; in a real case the satellite positions would notbe such simple vectors).

Table 1. Satellite data.

Satellite Position Time

1 (1, 2, 0) 19.92 (2, 0, 2) 2.43 (1, 1, 1) 32.64 (2, 1, 0) 19.9

Let (x, y, z) be the ship’s position and t the time when the signals arrive. Our goalis to determine the values of these variables. Using the data from the first satellite, wecan compute the distance from the ship as follows. The signal was sent at time 19.9and arrived at time t . Traveling at a speed of .047, that makes the distance

d = .047(t − 19.9).

This same distance can be expressed in terms of (x, y, z) and the satellite’s position(1, 2, 0):

d =√

(x − 1)2 + (y − 2)2 + (z − 0)2.

Combining these results leads to the equation

(x − 1)2 + (y − 2)2 + z2 = .0472(t − 19.9)2. (1)

Similarly, we can derive a corresponding equation for each of the other three satel-lites. That gives us four equations in four unknowns, and so we can solve for x, y, zand t. These are not linear equations, but we can use algebra to obtain a linear sys-tem that we can solve. Before proceeding, however, let us consider the geometry a bitmore.

Pure spatial triangulation has a simple geometric representation. Specifying the dis-tance from an unknown position P to a known fixed point F restricts P to lie on asphere centered at F . Given the distances to three points, F1, F2, F3, we conclude thatP must lie at a point of intersection of three spheres. The intersection is typically apair of points. To see this, observe that the intersection of any two of the spheres is acircle, which (except for degenerate cases) meets the third sphere in two points. More-over, there is a simple characterization of the points. Consider the plane that contains

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the intersection circle of the first two spheres. It contains the points where all threespheres meet. These points are similarly contained in the plane of intersection for thesecond and third spheres. The intersection of the two planes is a line, and the pointswe seek are the intersections of that line with any of the three spheres. Ultimately wehope to eliminate one of these intersection points. For GPS, knowing that the receiveris on the surface of the earth permits this final step.

For the GPS problem, triangulation takes place in both space and time, complicatingthe geometry somewhat. As a simplification, let us look at triangulation in a plane,rather than in space. A receiver at an unknown position P obtains simultaneous signalsfrom three sources. The locations of the sources and the times at which the signals werebroadcast are known. Locating P can be visualized dynamically. Imagine each signalradiating outward in a circular wave. There are three of these waves, initiated at threedifferent times, as if three pebbles were cast into a still pond. We are trying to locate thepoint at which the outermost ripples from each stone first meet. This view is capturedin several snapshots in Figure 1.

Figure 1. Ripples radiate from three stones.

An alternative to this dynamic visualization is obtained by considering a three di-mensional space-time. The horizontal plane represents spatial positions, and the ver-tical axis represents time. Each point (x, y, t) thus identifies a specific location andtime. In this setting, the snapshots of Figure 1 are horizontal sections of a three dimen-sional figure. For each signal there is a cone. A point (x, y, t) is on that cone if thesignal arrives at point (x, y) at time t . The vertex of the cone is at (x0, y0, t0), wheret0 indicates when the signal was broadcast, and (x0, y0) is the location from which thesignal was sent. Our problem is to find a point of intersection of three distinct lightcones.

As depicted in Figure 2, the intersection of two cones lies in a plane. Arguing asbefore, with three cones we can form two planes of intersection. These meet in a line

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Figure 2. Two intersecting cones.

which must contain the points common to all three cones. Intersecting the line withany one cone then determines the points we want.

This geometric picture is an exact analog of the GPS problem. Indeed, we may view(1) as a cone in four dimensional space-time. With four satellites there will be a systemof four similar equations, and the solutions are points of intersection of four lightcones. As in the three dimensional case, we can intersect the cones in pairs to identify(hyper)planes, and the intersection of three of these specifies a line. Algebraically, thiscorresponds to solving an underdetermined system of linear equations, and the line isthe general set of solutions. We will look at the details of the algebra next.

Algebraic Solution. Let us focus again on the equation for the first satellite:

(x − 1)2 + (y − 2)2 + z2 = .0472(t − 19.9)2.

Expanding all the squares and rearranging leads to this version:

2x + 4y − 2(.0472)(19.9)t = 12 + 22 − .0472(19.9)2 + x2 + y2 + z2 − .0472t2.

Similar equations can be derived for the three other satellites. Writing all four equa-tions together gives

2x + 4y + 0z − 2(.0472)(19.9)t = 12 + 22 + 02 − .0472(19.9)2

+ x2 + y2 + z2 − .0472t2

4x + 0y + 4z − 2(.0472)(2.4)t = 22 + 02 + 22 − .0472(2.4)2

+ x2 + y2 + z2 − .0472t2

2x + 2y + 2z − 2(.0472)(32.6)t = 12 + 12 + 12 − .0472(32.6)2

+ x2 + y2 + z2 − .0472t2

4x + 2y + 0z − 2(.0472)(19.9)t = 22 + 12 + 02 − .0472(19.9)2

+ x2 + y2 + z2 − .0472t2

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The quadratic terms in all the equations are the same, so by subtracting the firstequation from each of the other three, we obtain a system of three linear equations:

2x − 4y + 4z + 2(.0472)(17.5)t = 8 − 5 + .0472(19.92 − 2.42)

0x − 2y + 2z − 2(.0472)(12.7)t = 3 − 5 + .0472(19.92 − 32.62)

2x − 2y + 0z + 2(.0472)(0)t = 5 − 5 + .0472(19.92 − 19.92)

Geometrically, each of these equations represents a hyperplane containing the inter-section of two light cones.

Now we know that this system cannot have a unique solution. But if the satellitedata are accurate, there must be a solution to the original system of quadratic equations,and this linear system must be consistent. By deriving the general solution, it will bepossible to express three of the unknowns in terms of the fourth. Then, substitutionin one of the original quadratic equations will produce a quadratic equation in onevariable. Solving that will lead, in turn, to values for the other three variables.

So, proceeding according to this plan, we formulate the linear system as an aug-mented matrix:

2 −4 4 .077 3.860 −2 2 −.056 −3.472 −2 0 0 0

.

In this matrix, the integer entries should be considered to be exact values. The otherentries were computed to approximately sixteen-place accuracy (using a similarly ac-curate value for the speed of light), and then rounded to three decimal digits. This isa convenience for the sake of appearances that will be adhered to through all the fol-lowing calculations. In general, all of the computations are carried through with fullaccuracy, but appear with only a few decimal digits in print.

Continuing with the solution of the system, the reduced row echelon form for theaugmented matrix is

1 0 0 .095 5.410 1 0 .095 5.410 0 1 .067 3.67

.

Therefore, we have the general solution

x = 5.41 − .095t, y = 5.41 − .095t, z = 3.67 − .067t, t free.

Returning to (1), and substituting the above expressions for x, y, and z, we obtain

(5.41 − .095t − 1)2 + (5.41 − .095t − 2)2 + (3.67 − .067t)2 = .0472(t − 19.9)2

or

0.02t2 − 1.88t + 43.56 = 0,

leading to two solutions, 43.1 and 50.0. If we select the first solution, then (x, y, z) =(1.317, 1.317, 0.790), which has a length of about 2. We are using units of earth radii,so this point is around 4000 miles above the surface of the earth. The second value oft leads to (x, y, z) = (.667, .667, .332), with length 0.9997. That places the point onthe surface of the earth (to four decimal places) and gives us the location of the ship.

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Of course, to use this information, we would most likely want to convert it to a latitudeand longitude, but that computation will not be presented here.

To double check the accuracy of these results, they can be substituted in the originalfour quadratic equations. For this particular example, however, I concocted the orig-inal problem by starting with assumed values for the unknowns, (2/3, 2/3, 1/3) for(x, y, z) and 50 for t , as well as the positions for the satellites. Using these values Idetermined the length of time it would take the ship to receive a signal from each satel-lite. That is how I obtained the times in the last column of Table 1. Rounding thosetimes to one decimal place introduced some small errors, so that the computed solutionfor (x, y, z) is not quite equal to (2/3, 2/3, 1/3). Still, the computed results came outclose enough to the correct values to provide some confidence in the methods.

Using this approach, it is possible to create other examples for use in a linear algebraclass. You can place the satellites and the ship in any positions you wish. Of course,it is important that the satellites be above the horizon as viewed from the ship. (It isa nice exercise to show that a satellite at position s is above the horizon from a pointp on the unit sphere if and only if s · p > 1.) Alternatively, you can specify values forall of the data in Table 1, but then there is no guarantee that the GPS receiver is on thesurface of the earth.

Accuracy Estimates. The purpose of this example has been to present a realisticsituation in which finding the general solution of an underdetermined system playsan important role. As noted earlier, GPS receivers do not operate along the lines ofthe example. While the example is therefore not an accurate model of computationsperformed by GPS receivers, it nevertheless serves as a practical conceptual tool foranalyzing GPS performance. For instance, the example gives insight about the processof spatial triangulation, and the general framework for GPS. It is the sort of ideal-ized computation that one might want to make to get a feel for the relationships, andwhile there are many ways to solve the original system of quadratic equations, the onepresented above is certainly efficient and practical.

In addition, the simple geometric model provides a context in which other aspects ofthe system can be studied. One important example concerns the relationship betweenuncertainties in the measured data (satellite positions and times) and the resultant un-certainties in the GPS results. It turns out that this relationship is dependent on thegeometry of the satellites and the GPS receiver [1].

As an illustration of this idea, consider again starting with hypothesized positionsfor the satellites as well as the position and time data for the ship, and computingthe appropriate time data for the satellites. This gives us a consistent set of true data.Now to simulate the effects of measurement error, add a random perturbation to eachof the satellite data. This gives us a perturbed version of Table 1. Based on theseperturbed values, solve the equations to compute a predicted position and time for theship. Comparing these computed values with the prespecified true values shows theeffects of the simulated errors in the satellite data. For the example above, adding errorson the order of .001 (about 4 miles spatially and 10−6 seconds in time) introduces errorson the order of 5 miles in the computed position of the ship. Similarly, perturbationsof around 10−6 in the satellite data causes errors of around 10 yards for the computedposition of the ship.

Let’s consider a different geometric configuration, by changing the location of thefourth satellite to (1.1, .9, 1.2), fairly close to satellite 3. This time, perturbing thesatellite data by about .001 results in errors around 50 miles in the computed shipposition—10 times more than in the first case. A similar tenfold increase in error isobserved when the satellite data are perturbed by about 10−6. This illustrates that the

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impact of errors in the satellite data depends strongly on the geometric configuration.It also shows how computations in the conceptual geometric model contribute to anunderstanding of the more complicated real system.

This final discussion would probably not be appropriate in most linear algebraclasses. It would require too great a digression from the main ideas of the course,and is not really necessary for understanding the use of linear algebra in the simplifiedgeometric model presented above. Indeed, the computation of an unknown position inthat model is the primary emphasis of this paper, as a simple application of solvingunderdetermined linear systems. On the other hand, it is worthwhile for the instructorto know the context in which applications are really used. In this example, it would beincorrect to suggest to students that GPS receivers use something like the calculationsabove to determine positions. However, the simple geometric model does have value,for example in modeling performance characteristics of the real system.

References

1. Paul Massatt and Karl Rudnick, Geometric formulas for dilution of precision formulas, Journal of Navigation37:4 (1990) 379–392.

2. Gilbert Strang, The mathematics of GPS, SIAM News 30:5 (1997). Also online at http://www.siam.org/siamnews/general/gps.htm

3. Gilbert Strang and Kai Borre, Linear Algebra, Geodesy, and GPS, Wellesley-Cambridge Press, 1997.4. Richard B. Thompson, Global positioning system: the mathematics of GPS receivers. Mathematics Magazine

71:4 (1998) 260–269.

The Bipolar Oven

Marc Brodie, of the College of St. Benedict (mbrodie@csbsju.edu), who keepshis eyes open, noticed a sign at a pizza place:

THE BESTPIZZA IS

A MATTER OFDEGREES

CAUTION! OVEN +/ − 500◦

He comments, “While I concede that both temperatures would warrant cau-tion (and ignoring the fact that −500◦ is below absolute zero), only one willdo a reasonable job of cooking a pizza. (Incidentally, this sign was obviously amass-produced-by-headquarters type sign, not a handwritten scribble on a post-itnote.)”

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